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topics_11_3_11_4_11_5

# topics_11_3_11_4_11_5 - Topics 11.3 11.4 11.5 11.3 Polar...

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Topics 11.3, 11.4, 11.5 11.3 Polar Coordinates Polar coordinate system introduced by Newton : It has a pole that is the origin O of a Cartesian coordinate system . It has a polar axis that coincides with the right-half of the x -axis in the Cartesian coordinate system, starting from the pole O. If P 6 = O is a point on the plane and r is the distance from the pole O to the point P and θ is the angle between the polar axis and the line OP assuming that the counter-clockwise direction is a positive direction, then ( r , θ ) are the polar coordinates of the point P . We agree the pole O to be represented by (0 , θ ) for any value of θ ( r = 0 at the pole O ). Now suppose that the point P is tracing a curve C . Then r = f ( θ ) , α θ β is a representation of the curve C called polar representation. In fact, it is equivalent to the following parametric equations of the curve C : x = f ( θ ) cos( θ ) , y = f ( θ ) sin( θ ) , α θ β. The equation r = f ( θ ) , α θ β is called a polar equation of a curve. It is tracing a curve. The angle θ is a parameter in the polar representation of a curve. 1

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Important remarks for plots! (1) We extend the meaning of polar coordinates ( r, θ ) to r negative, agreeing that the points ( - r , θ ) and ( r , θ ) are on the same line, equidistant to O , but opposite with respect to O . In other words, ( - r , θ ) represents a point with polar coordi- nates ( r , θ + π ) . (2) A polar curve (a curve given by a polar equation) is symmetric about the x - axis if r ( - θ ) = r ( θ ). (3) A polar curve is symmetric about the y - axis if r ( π - θ ) = r ( θ ). (4) A polar curve is symmetric with respect to the origin if r ( π + θ ) = r ( θ ), i.e., if replacing r by - r we obtain an equivalent polar equation. Problem 1. Sketch the region in the plane consisting of points whose polar coordinates satisfy 1 r < 3 , - π/ 3 < θ 2 π/ 3 . Problems 31, 35, 36, 37, 41, 42/ page 684. Sketch the curve with the given polar equation. 31 . r = sin( θ ); 35 . r = θ, θ 0; 36 . r = ln( θ ) , θ 1 37 . r = 4 sin(3 θ ); 41 . r = 1 - 2 sin( θ ); 42 . r = 2 + sin( θ ) 2
By using CAS MAPLE code plot ([[ 1 * cos ( t ) , 1 * sin ( t ) , t = - Pi / 3 .. 2 * Pi / 3 ] , [ 3 * cos ( t ) , 3 * sin ( t ) , t = - Pi / 3 .. 2 * Pi / 3 ] , [ r * cos ( - Pi / 3 ) , r * sin ( - Pi / 3 ) , r = 1 .. 3 ] , [ r * cos ( 2 * Pi / 3 ) , r * sin ( 2 * Pi / 3 ) , r = 1 .. 3 ]] , color = [ black , red , red , black ]); we plot the region to verify the result of Problem 1. 3

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Plots of Problems 31, 35, 36 by using CAS MAPLE polar- plot. We give an explicit form of the CAS MAPLE polar-plot for Problem 31: r = sin( θ ). For convenience we use the notation t instead of θ : polarplot ( sin ( t ) , t = 0 .. 2 * Pi ); Plots of Problems 37, 41, 42 by using CAS MAPLE polar-plot. 4
Problems 15-20, page 684. Identify the curve by finding a Cartesian equation for the curve: 15 . r = 2; 16 . r cos( θ ) = 1; 17 . r = 3 sin( θ ); 18 . r = 2 sin( θ ) + 2 cos( θ ); 19 . r = csc( θ ); 20 . r = tan( θ ) sec( θ ) . Hint. Use x = r cos( θ ) , y = r sin( θ ). Solution of Problem 18. Multiplying both sides by r we get r 2 = 2 r sin( θ )+2 r cos( θ ) x 2 + y 2 = 2 y +2 x ( x - 1) 2 +( y - 1) 2 = 2 and this is a circle of radius 2, centered at (1 , 1).

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